Chem

Suppose that you want to construct a galvanic cell from nickel and cadmium. You use 0.01M sol of cadmium nitrate as one of the electrolytes. If you need a cell potential of 0.17V for this system, what concentration of NiCl2 should you use?

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asked by Keena
  1. The cell reaction will be
    Cd(s) + Ni^2+(aq) ==> Ni(s) + Cd(0.01M)

    step 1. Calculate the E value for the Cd part using
    E = Eo - (0.05916/n)*log (1/0.01) and the equation for the reduction of Cd^2+ to Cd(s) as in Cd^2+ + 2e ==> Cd(s). Now reverse the sign since you want the oxidation of Cd and not the reduction of Cd^2+.

    Step 2. The half cells for the first reaction I gave above are
    Cd(s)==> Cd^2+ + 2e .........E1 = from step 1
    Ni^2+ + 2e ==> Ni(s)............E 2= ?
    ---------------------------------------------
    Add to get Cd + Ni^2+ =>Cd^2+ + Ni E3 = 0.17 v.
    Now calculate E2 = ? from E1 + E2 = E3

    Step 3. Now you calculate the concentration of Ni^2+ from
    Ni^2+ + 2e ==> Ni and E = Eo - (0.05916/n)*log[1/(Ni)].
    Plug in Eo for Ni^2+ reduction, Eo is from the reduction tables, and E is the calculation from step 2 of E2. Solve for (Ni^2+).

    Post ALL of your work if you get stuck.

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